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    <div class="post-body" itemprop="articleBody"><h1 id="基本递推">基本递推</h1>
<p>递推是编程中最常用的技能，规模较大的问题可以由规模较小的问题计算得到，从规模较小的问题开始，按一定的顺序推导更大问题的解的过程。</p>
<span id="more"></span>
<h3 id="例斐波那契数列">例：斐波那契数列</h3>
<p><span class="math inline">\(f_{1}=1, f_{2}=1,
f_{n}=f_{n-1}+f_{n-2}\)</span>.</p>
<p>求数列的第 <span class="math inline">\(n\)</span> 项.</p>
<p>当这类问题存在多次提问时，可以用一个数组把序列的表提前算好，俗称“打表”.</p>
<blockquote>
<p>知识点：有的问题（或函数，或数列）随着自变量的增大，增长非常快，这样稍大一些的时候，<code>long long</code>
都存不下。出题人为了考察这类问题，会要求输出答案对一个较大的质数取模的结果，这个结果正确，就认为答案正确。</p>
</blockquote>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 斐波那契数列前 1000 项，每项都是对 10^9+7 取模的结果</span></span><br><span class="line"><span class="type">const</span> <span class="type">int</span> mod = <span class="number">1e9</span> + <span class="number">7</span>;</span><br><span class="line"><span class="type">long</span> <span class="type">long</span> f[<span class="number">1111</span>];</span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>&#123;</span><br><span class="line">    f[<span class="number">1</span>] = f[<span class="number">2</span>] = <span class="number">1</span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">3</span>; i &lt;= <span class="number">1000</span>; i ++) &#123;</span><br><span class="line">        f[i] = (f[i - <span class="number">1</span>] + f[i - <span class="number">2</span>]) % mod;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="type">int</span> n;  <span class="comment">// 不断询问的第“n”项</span></span><br><span class="line">    <span class="keyword">while</span>(<span class="built_in">scanf</span>(<span class="string">&quot;%d&quot;</span>, &amp;n) != EOF) &#123; <span class="comment">// EOF是scanf读到输入文件末尾时的返回值</span></span><br><span class="line">        <span class="built_in">printf</span>(<span class="string">&quot;%d\n&quot;</span>, f[n]);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h3 id="例传染">例：传染</h3>
<p>数量不限个健康动物，一个生病动物每轮会传染 <span
class="math inline">\(x\)</span> 个健康动物， 求 <span
class="math inline">\(n\)</span> 轮后染病动物总数。</p>
<p>输入 <span class="math inline">\(x\)</span> 和 <span
class="math inline">\(n\)</span></p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">10 2</span><br></pre></td></tr></table></figure>
<p>输出</p>
<figure class="highlight txt"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">121</span><br></pre></td></tr></table></figure>
<p>每轮结束时，染病动物总数应当是传染的数量加上已染病的数量，形成递推关系。</p>
<p>如果涉及多次查询，就开数组大表，否则用一个变量递推就行。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstdio&gt;</span></span></span><br><span class="line"><span class="meta">#<span class="keyword">include</span><span class="string">&lt;cstring&gt;</span></span></span><br><span class="line"><span class="function"><span class="type">int</span> <span class="title">main</span><span class="params">()</span> </span>&#123;</span><br><span class="line">    <span class="type">int</span> x, n;</span><br><span class="line">    <span class="type">long</span> <span class="type">long</span> ans = <span class="number">1</span>;          <span class="comment">// 最开始的一个，指数级增长要注意查看题目数据范围</span></span><br><span class="line">    <span class="built_in">scanf</span>(<span class="string">&quot;%d%d&quot;</span>, &amp;x, &amp;n);            </span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">1</span>; i &lt;= n; i ++) &#123;</span><br><span class="line">        ans = ans * x + ans;    <span class="comment">// 传染的加上已有的</span></span><br><span class="line">    &#125;</span><br><span class="line">    <span class="built_in">printf</span>(<span class="string">&quot;%lld\n&quot;</span>, ans);</span><br><span class="line">    <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="常见注意点小结">常见注意点小结</h2>
<ul>
<li>注意递推问题的增长规模与数据范围，选择性使用
<code>long long</code></li>
<li>题目答案要求对特定数取模，要注意不仅仅是结果取模，要避免计算过程中超类型范围
<ul>
<li>递推式多项累加，要分析中间结果是否会超<code>long long</code>，如果会，则要随时取模：
<code>f[n] = ((f[n - 1] + f[n - 2]) % mod + f[n - 3]) % mod</code></li>
<li>递推式带有乘法，虽然取模数不超 <code>int</code> ，但乘法计算超：
用“1LL”去乘中间结果临时转 <code>long long</code>：
<code>f[n] = 1LL * f[n-1] * f[n - 2] % mod</code>;</li>
</ul></li>
</ul>

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